Consider the initial value problem y\" + 2y\' + y = 2e^-t,y(0) = a, y\'(0) = b|
ID: 3402334 • Letter: C
Question
Consider the initial value problem y" + 2y' + y = 2e^-t,y(0) = a, y'(0) = b| where a and b are constants. The DE describes a system with damped motion. But the specific behavior will depend on a and b. Some solutions for different values of a and b are shown below: Determine the solution y(t) of this IVP. Show that irrespective of the values of a and b. Now assume that b = - 1. What are the conditions on a so that the solution y(t) has no local extreme on the interval [0, infinity) has exactly one local minimum and one local maximum on the interval [0, infinity) has one local minimum or maximum on the interval [0, infinity) Is it possible to find a value for a so that y(t) has local extreme at both t = 1 and t = 2?Explanation / Answer
The characteristic equation has repeated roots -1,-1.
So the complimentary solution is
A e-t+Bte-t
A particular solutin is given by
t2e-t
So a general solution is
y(t) =A e-t+Bte-t+ t2e-t
The initial conditions are y(0)=a and y'(0)=b.
(a) Plugging in these conditions , the solution of the initial value problem is
y(t) =a e-t+(a+b)te-t+ t2e-t ......................................(1)
(b) as e-t tends to 0 as t tends to infinity, y(t) tends to 0 irrespective of the values of a and b.
(c) (i) if b =-1, then
y(t) = a e-t+(a-1)te-t+ t2e-t .
y'(t) = -ae-t + (a-1) [-te-t +e-t ] +2te-t -t2 e-t
=-e-t [ t2 +(a-3)+1]
(a) no local extrema if discriminant= (a-3)2 -4 >0 (no real roots)
(b) if the discriminant is equal to 0
(c) if the discriminant is positive
(d) As the product of the roots is 1 , it is not possible to find such a value for a.